Answer
$x \approx 4.405$
Work Step by Step
$\log_3 (x+2) + \log_3 (x-3) = 2$
$\log_3 (x+2)(x-3) = 2$
$3^{2} = (x+2)(x-3)$
$9 = (x+2)(x-3)$
$9 = x(x-3)+2(x-3)$
$9 = x^{2} - 3x + 2x - 6$
$x^{2} -x - 6 - 9 = 0$
$x^{2} - x - 15 = 0$
$x = \frac{-(-1)±\sqrt {(-1)^{2}-4(1)(-15)}}{2(1)}$
$x = \frac{1±\sqrt {1-4(1)(-15)}}{2}$
$x = \frac{1±\sqrt {1+60}}{2}$
$x = \frac{1±\sqrt {61}}{2}$
$x \approx 4.405, -3.405$
Since $x$ can't be negative, the only possible solution is $x = 4.405$. This is because $\log$ of a negative number does not exist and therefore we only accept the positive number of $x$.
Check:
$\log_3 ((\frac{1+\sqrt {61}}{2})+2) + \log_3 ((\frac{1+\sqrt {61}}{2})-3) \overset{?}{=} 2$
$\log_3 (6.405124...) + \log_3 (1.405124...) \overset{?}{=} 2$
$\log_3 [(6.405124...) \times (1.405124...)] \overset{?}{=} 2$
$\log_3 9\overset{?}{=} 2$
$\log_3 3^{2}\overset{?}{=} 2$
$2\log_3 3\overset{?}{=} 2$
$2(1)\overset{?}{=} 2$
$2 = 2$
